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I am reading a proof in Baby Rudin that perfect sets in $\mathbb R^k$ are uncountable. The proof seems to assume it is first countable and show that it contradicts the fact that if a collection of compact sets have non empty finite intersections then the entire intersection is nonempty.

However, is it possible for an open neighborhood of a point in an uncountable set, say $x_0,$ to intersect only a countably infinite number of points? Because then it seems as if the same proof can be applied to this subset only.

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  • $\begingroup$ Open neighborhood of a point in an uncountable set or in a perfect set? $\endgroup$ – Adayah Aug 18 '17 at 16:32
  • $\begingroup$ What is "it" in "it is first countable." $\endgroup$ – Thomas Andrews Aug 18 '17 at 16:32
  • $\begingroup$ It is possible for the intersection of an open subset of $\mathbb{R}^k$ and an uncountable subsets of $\mathbb{R}^k$ to be countable---consider $K = [0,1]\cup\{3\}$ and $U = (2,4)$. However, $K$ has an isolated point, and is therefore not perfect. $\endgroup$ – Xander Henderson Aug 18 '17 at 16:33
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    $\begingroup$ Well, you don't assume it is first-countable. If $X$ is first countable, and $Y\subseteq X$ has the induced topology, then $Y$ is also first countable. $\endgroup$ – Thomas Andrews Aug 18 '17 at 16:37
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    $\begingroup$ To restate my question, say we have a perfect set $P$ that is uncountable. Can we have a neighborhood of a point in $P$ intersect $P$ at only a countably infinite number of points? Sorry if I wasn't too clear. $\endgroup$ – 伽罗瓦 Aug 18 '17 at 16:42
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Consider the set $\{0\}\cup\{1/n : n \in \mathbb{N}\} \cup\{x :1 \leq x\leq 2, x \in \mathbb{R}\}$ which is uncountable and any ball of radius less than one around zero will only have countably many points and is an open neighborhood. Notice that it has isolated points so it is not perfect.

Also consider $\{x : 0 \leq x \leq 1, x \in \mathbb{Q}\} \cup\{x : 1 \leq x \leq 2, x \in \mathbb{R}\}$ which is uncountable, any of the rational numbers on the interval $[0,1]$ can have a ball of sufficiently small radius to contain only a countable number of points and it has no isolated points, but it is not closed so it is not perfect.

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    $\begingroup$ Ah I see. Thank you for your examples. So is it not possible if the set were perfect? $\endgroup$ – 伽罗瓦 Aug 18 '17 at 16:44
  • $\begingroup$ @ntntnt That's correct. A perfect set is equal to its derived set so every point must be a limit point. This implies every neighborhood around any point must be non-empty. If a neighbor in a set is countable and not dense on that region in $\mathbb{R}$ it will have an isolated point. If it's countable and dense then not all the points in that neighborhood are limit points of the set. If a set is perfect neither of these cases can be true. $\endgroup$ – CyclotomicField Aug 18 '17 at 17:03
  • $\begingroup$ Can you explain why if a neighborhood of a point in a set is countable and not dense on that region in $R$ it will have an isolated point? I don't see why this is true. If there is a point in the region that isn't a limit point of our neighborhood, does this mean the point is isolated? $\endgroup$ – 伽罗瓦 Aug 18 '17 at 17:15
  • $\begingroup$ @ntntnt a dense subset $A$ of the reals implies that a point in the reals is either in $A$ or a limit point of $A$. Since it's not dense this implies there exists a point in $R$ which is neither in $A$ nor a limit point of $A$. So if $x$ is a point with this property there exists a neighborhood around that point which only contains $x$, thus it's isolated. $\endgroup$ – CyclotomicField Aug 18 '17 at 17:22
  • $\begingroup$ Sorry, doesn't that mean that there exists a neighborhood around $x$ that doesn't intersect $A$ at all, not necessarily isolated? You say that " If a neighbor in a set is countable and not dense on that region in $\mathbb{R}$ it will have an isolated point." That just means there is a point $x$ that is neither in our neighborhood nor a limit point of that neighborhood. But why should it be an isolated point of our perfect set? $\endgroup$ – 伽罗瓦 Aug 19 '17 at 21:50

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